Question

use factor theorem to solve this 2x^4+x^2-1

Answer 1

are you sure there isn't a typo in the question...

Answer 2

ok sorry just solve the equation however :)

Answer 3

let u = x^2 then 2u^2 + u - 1 = 0 (2u-1)(u +1)=0

then u = 1/2 or -1 resubstituting

x^2 = 1/2 and -1

\[solutions are x = 1/\sqrt{2} \] and x = i

Answer 4

oops should be \[x = \pm 1/\sqrt{2} , \pm i\]

Answer 5

wasn't this question already answered today (or yesterday depending on where you live)

Answer 6

Should only be 4 zeros here, not 5

Answer 7

Descartes rule of signs
f(x)=2x^4+x^2-1
Should be exactly 1 real positive zero
f(-x)=2x^4+x^2-1
Should be exactly 1 real negative zero
This leaves 2 complex conjugates for the other two zeros.
There are no rational zeros given by the rational zeros test.
Graphing on a calculator, it looks like sqrt(2)/2 will work

Synthetic division time
sqrt(2)/2| 2 0 1 0 -1
sqrt(2) 1 sqrt(2) 1
2 sqrt(2) 2 sqrt(2) | 0
Q(x)=2x^3+sqrt(2)x^2+2x+sqrt(2)
Since there were rational coefficients in the original polynomial, irrational zeros occur in conjugate pairs, meaning -sqrt(2)/2 is also a 0. We will test this with synthetic division on the previous quotient (Q(x))
-sqrt(2)/2| 2 sqrt(x) 2 sqrt(2)
-sqrt(x) 0 -sqrt(2)
2 0 2 | 0
So it's obvious from here that sqrt(2)/2 and -sqrt(2)/2 are zeros.
Q(x)=2x^2+2
We are now left with a quadratic equation which we will use to solve the other zeros. Using descartes rule of signs above and knowing that if the real numbers comprise the coefficients of the polynomial, we know both answers should be complex conjugates.
2x^2+2=0
x^2+1=0
x^2=-1
x= sqrt(-1), -sqrt(-1)
x=i, -i
So, overall, x=sqrt(2)/2, -sqrt(2)/2, i, -i
We can write a factored form of f(x) and prove these are the right answers by expanding the factored form back into the expanded form:
(x-sqrt(2)/2)(x+sqrt(2)/2)(x-i)(x+i)
(x^2-1/2)(x^2+1)
(x^4+x^2-(1/2)x^2-1/2
x^4+(1/2)x^2-1/2
This above equation is the simplest form of the equation, we still need to multiply it by a constant to get it exactly the same way it was written. Obviously we'll use 2 for this constant.
2x^4+x^2-1

Answer 8

(x^2+1)(2x^2-1)

Answer 9

wow.... a simply substitution achieved the the same as Descartes.... 4 roots, 2 real and 2 complex.... and

\[1/\sqrt{2} = \sqrt{2}/2\] the denominator has been rationalised